tìm các giá trị của a để \(lim\left(\left[\sqrt{a-1}-3\right]n^2+5n-3\right)=-\infty\)
Giá trị của các giới hạn :
a, lim\(\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+1}\right)\) khi x→\(-\infty\)
b, lim\(\left(\sqrt{x^2+x}-\sqrt[3]{x^3-x^2}\right)\) khi x→\(+\infty\)
c, lim\(\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\) khi x→\(+\infty\)
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)
Check lai ho minh nhe :v
tìm giới hạn của dãy số
1.\(\lim\limits_{n->\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
2.\(\lim\limits_{n->\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
3.tìm a,b để \(\lim\limits_{n->\infty}\left(\sqrt{an^2+bn+2}-2n\right)=2\)
1: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^3+n^2+n+1-n^3}{\sqrt[3]{\left(n^3+n^2+n+1\right)^2}+n\cdot\sqrt[3]{n^3+n^2+n+1}+n^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+1}\)
\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)
2: \(\lim\limits_{n\rightarrow\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}\)
\(=\dfrac{2}{1+1}=\dfrac{2}{2}=1\)
bài 1
a. \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+3n+1}-n\right)\)
b. \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+2n}-\sqrt{n^2+1}\right)\)
c.\(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+n}-\sqrt[3]{n^3-2n^2}\right)\)
bài 2
a. \(\lim\limits_{n\rightarrow+\infty}\left(2n-\sqrt{n^2+3n}\right)\)
b. \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+2}-\sqrt{3n+1}\right)\)
c. \(\lim\limits_{n\rightarrow+\infty}\left(n\sin2n-3n^3\right)\)
Giới hạn của dãy nên bạn tự hiểu n tiến tới dương vô cực
1.
\(lim\frac{3n+1}{\sqrt[3]{\left(n^3+3n+1\right)^2}+n\sqrt{n^3+3n+1}+n^2}=lim\frac{3+\frac{1}{n}}{\sqrt[3]{\frac{\left(n^3+3n+1\right)^2}{n^3}}+\sqrt{n^3+3n+1}+n}=\frac{3}{\infty}=0\)
b=\(lim\left(\sqrt[3]{n^3+2n}-n+n-\sqrt{n^2+1}\right)=lim\left(\frac{2n}{\sqrt[3]{\left(n^3+2n\right)^2}+n\sqrt[3]{n^3+2n}+n^2}-\frac{1}{n+\sqrt{n^2+1}}\right)\)
\(=lim\left(\frac{2}{\sqrt[3]{\frac{\left(n^3+2n\right)^2}{n^3}}+\sqrt[3]{n^3+2n}+n}-\frac{1}{n+\sqrt{n^2+1}}\right)=0-0=0\)
c\(=lim\left(\frac{2n^2+n}{\sqrt[3]{\left(n^3+n\right)^2}+\sqrt[3]{\left(n^3+n\right)\left(n^3-2n^2\right)}+\sqrt[3]{\left(n^3-2n^2\right)^2}}\right)\)
\(=lim\left(\frac{2+\frac{1}{n}}{\sqrt[3]{\left(1+\frac{1}{n^2}\right)^2}+\sqrt[3]{\left(1+\frac{1}{n^2}\right)\left(1-\frac{2}{n}\right)}+\sqrt[3]{\left(1-\frac{2}{n}\right)^2}}\right)=\frac{2}{1+1.1+1}=\frac{2}{3}\)
2.
a\(=lim\left[n\left(2-\sqrt{1+\frac{3}{n}}\right)\right]=+\infty\left(2-1\right)=+\infty\)
\(b=lim\left[n\left(\sqrt{1+\frac{2}{n^2}}-\sqrt{\frac{3}{n}+\frac{1}{n^2}}\right)\right]=+\infty\left(1-0\right)=+\infty\)
\(c=lim\left[n^3\left(\frac{sin2n}{n^2}-3\right)\right]=+\infty\left(0-3\right)=-\infty\)
Tính các giới hạn
a) \(\lim\limits_{x\rightarrow+\infty}\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\)
b) \(\lim\limits_{x\rightarrow+\infty}\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\)
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[n]{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)}-x\right)\\ =\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)-x^n}{\sqrt[n]{\left(\left(x+a_1\right)\left(x+a_2\right)...\left(x+a_n\right)\right)^{n-1}}+...+x^{n-1}}\right)\)
= hệ số xn-1 trên tử/hệ số xn-1 dưới mẫu = \(\dfrac{a_1+a_2+...+a_n}{n}\)
a. Có bao nhiêu giá trị của a để \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-ax+2021}-x+1\right)=a^2\)
b. Tìm a để hàm số f(x)=\(\left\{{}\begin{matrix}\dfrac{x^3+1}{x+1}khix\ne-1\\3akhix=-1\end{matrix}\right.\)gián đoạn tại điểm \(x_0=-1\)
c. Cho tứ diện đều ABCD .Góc giữa 2 vecto DA và BD bằng?
d. Cho hàm số y = f(x) = \(\dfrac{x^2-1}{2-2x}\)khi \(x\ne1\) .Để hàm số liên tục tại x=1 thì f(1) phải nhận giá trị nào dưới đây? (giải tự luận giúp em ạ)
A.-1 B.1 C.2 D.0
e. Cho hàm số \(f\left(x\right)=x^3+2x-1\) .Xét phương trình f(x) = 0 (1), trong các mệnh đề sau tìm mệnh đề sai? giải tự luận giúp em ạ
A. (1) có nghiệm rên khoảng (-1;1)
B. (1) Không có nghiệm trên khoảng (-5;3)
C. (1) có nghiệm trên R
D. (1) có nghiệm trên khoảng (0;1)
a.
\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2-ax+2021}-x+1\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(\sqrt{x^2-ax+2021}-x\right)\left(\sqrt{x^2-ax+2021}+x\right)}{\sqrt{x^2-ax+2021}+x}+1\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-ax+2021}{\sqrt{x^2-ax+2021}+x}+1\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{x\left(-a+\dfrac{2021}{x}\right)}{x\left(\sqrt{1-\dfrac{a}{x}+\dfrac{2021}{x^2}}+1\right)}+1\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{-a+\dfrac{2021}{x}}{\sqrt{1-\dfrac{a}{x}+\dfrac{2021}{x^2}}+1}+1\right)\)
\(=\dfrac{-a+0}{\sqrt{1+0+0}+1}+1=-\dfrac{a}{2}+1\)
\(\Rightarrow a^2=-\dfrac{a}{2}+1\Rightarrow2a^2+a-2=0\)
Pt trên có 2 nghiệm pb nên có 2 giá trị a thỏa mãn
b.
\(\lim\limits_{x\rightarrow-1}f\left(x\right)=\lim\limits_{x\rightarrow-1}\dfrac{x^3+1}{x+1}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}=\lim\limits_{x\rightarrow-1}\left(x^2-x+1\right)\)
\(=1+1+1=3\)
\(f\left(-1\right)=3a\)
Hàm gián đoạn tại điểm \(x_0=-1\) khi:
\(\lim\limits_{x\rightarrow-1}f\left(x\right)\ne f\left(-1\right)\Rightarrow3\ne3a\)
\(\Rightarrow a\ne1\)
c.
Tứ diện ABCD đều \(\Rightarrow\Delta ABD\) đều
\(\widehat{\left(\overrightarrow{DA};BD\right)}=180^0-\widehat{\left(\overrightarrow{DA};\overrightarrow{DB}\right)}=180^0-\widehat{ADB}=180^0-60^0=120^0\)
d.
\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{x^2-1}{2-2x}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x+1\right)}{-2\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{x+1}{-2}=\dfrac{1+1}{-2}=-1\)
Để hàm liên tục tại \(x=1\)
\(\Rightarrow f\left(1\right)=\lim\limits_{x\rightarrow1}f\left(x\right)=-1\)
e.
Hàm \(f\left(x\right)\) là hàm đa thức nên liên tục trên R
\(f\left(0\right)=-1< 0\) ; \(f\left(1\right)=2>0\)
\(\Rightarrow f\left(0\right).f\left(1\right)< 0\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;1\right)\)
Do \(\left(0;1\right)\) đồng thời là tập con của \(\left(-1;1\right)\) ; \(\left(-5;3\right)\) và R nên \(f\left(x\right)\) cũng có nghiệm trên các khoảng này
Vậy B là đáp án sai
có bao nhiêu giá trị nguyên của a có 1 chữ số để \(lim\left(\left(a^2-1\right)n^4+2n+11\right)=+\infty\)
Giới hạn đã cho bằng \(+\infty\)
\(\Leftrightarrow a^2-1\ge0\Rightarrow\left[{}\begin{matrix}a\ge1\\a\le-1\end{matrix}\right.\)
Có vô số giá trị nguyên
tính
a.\(\lim\limits_{n->+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
b.\(\lim\limits_{n->+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
c.\(\lim\limits_{n->+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
d.\(\lim\limits_{n->+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
a: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n^5+n^2-n+2}{\left(2n^3-1\right)\left(n^2+n+1\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(\dfrac{2n^3}{n^3}-\dfrac{1}{n^3}\right)\left(\dfrac{n^2+n+1}{n^2}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n^3}-\dfrac{1}{n^4}+\dfrac{2}{n^5}}{\left(2-\dfrac{1}{n^3}\right)\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}\right)}\)
\(=\dfrac{1}{2\cdot1}=\dfrac{1}{2}\)
b: \(\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{n^2-n+2}}{n+2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{n\left(1+\dfrac{2}{n}\right)}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\sqrt{1-\dfrac{1}{n}+\dfrac{2}{n^2}}}{1+\dfrac{2}{n}}=\dfrac{\sqrt{1-0+0}}{1+0}=\dfrac{1}{1}=1\)
c: \(\lim\limits_{n\rightarrow+\infty}\dfrac{n-\sqrt[3]{n^2-n^3}}{n^2+n+1}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{n}{n^2}-\dfrac{\sqrt[3]{n^2-n^3}}{n^2}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{\dfrac{1}{n}-\sqrt[3]{\dfrac{1}{n^4}-\dfrac{1}{n^3}}}{1+\dfrac{1}{n}+\dfrac{1}{n^2}}=\dfrac{0}{1}=0\)
d: \(\lim\limits_{n\rightarrow+\infty}\left(n-\sqrt{n^2+n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2-n^2-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-n-1}{n+\sqrt{n^2+n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{-1-\dfrac{1}{n}}{1+\sqrt{1+\dfrac{1}{n}+\dfrac{1}{n^2}}}=-\dfrac{1}{1+1}=-\dfrac{1}{2}\)
Tính các giới hạn sau
1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)
2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)
b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)
c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
tính
a. \(\lim\limits_{n->+\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
b.\(\lim\limits_{n->+\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
\(\lim\limits_{n\rightarrow+\infty}\left(\sqrt[3]{n^3+n^2+n+1}-n\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^3+n^2+n+1-n^3}{\sqrt[3]{\left(n^3+n^2+n+1\right)^2}+n\cdot\sqrt[3]{n^3+n^2+n+1}+n}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2+n+1}{\sqrt[3]{\left[n^3\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)\right]^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2+n+1}{n^2\cdot\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2}+n^2\cdot\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+n^2}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{1+\dfrac{1}{n}+\dfrac{1}{n^2}}{\sqrt[3]{\left(1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}\right)^2+\sqrt[3]{1+\dfrac{1}{n}+\dfrac{1}{n^2}+\dfrac{1}{n^3}}+1}}\)
\(=\dfrac{1}{1+1+1}=\dfrac{1}{3}\)
b: \(\lim\limits_{n\rightarrow+\infty}\left(\sqrt{n^2+n}-\sqrt{n^2-n+1}\right)\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{n^2+n-n^2+n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{2n-1}{\sqrt{n^2+n}+\sqrt{n^2-n+1}}\)
\(=\lim\limits_{n\rightarrow+\infty}\dfrac{2-\dfrac{1}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1-\dfrac{1}{n}+\dfrac{1}{n^2}}}=\dfrac{2}{\sqrt{1}+\sqrt{1}}=1\)